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3 years ago

5i / 3-4i division of the following

Mathematics

1 answer:

IceJOKER [234]3 years ago

7 0

Answer:

$\frac{5i}{3 - 4i} = \frac{3i - 4}{5}$

Step-by-step explanation:

Given

$\frac{5i}{3 - 4i}$

Required

Solve

We have:

$\frac{5i}{3 - 4i}$

Rationalize

$\frac{5i}{3 - 4i} = \frac{5i}{3 - 4i} * \frac{3 + 4i}{3 + 4i}$

$\frac{5i}{3 - 4i} = \frac{5i(3 + 4i)}{(3 - 4i)(3 + 4i)}$

Apply difference of two squares on the denominator

$\frac{5i}{3 - 4i} = \frac{5i(3 + 4i)}{3^2 - (4i)^2}$

$\frac{5i}{3 - 4i} = \frac{5i(3 + 4i)}{9 - (16*-1)}$

$\frac{5i}{3 - 4i} = \frac{5i(3 + 4i)}{9 +16}$

$\frac{5i}{3 - 4i} = \frac{5i(3 + 4i)}{25}$

Divide common factor (5)

$\frac{5i}{3 - 4i} = \frac{i(3 + 4i)}{5}$

Expand the numerator

$\frac{5i}{3 - 4i} = \frac{3i + 4*-1}{5}$

$\frac{5i}{3 - 4i} = \frac{3i - 4}{5}$

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<u>ANSWER:</u>

The midpoint of AB is M(-5,1). The coordinates of B are (-6, 7)

<u>SOLUTION: </u>

Given, the midpoint of AB is M(-5,1).

The coordinates of A are (-4,-5),

We need to find the coordinates of B.

We know that, mid-point formula for two points A $(x_{1}, y_{1})$ and B $(x_{1}, y_{2})$ is given by

$M\left(x_{3}, y_{3}\right)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

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Now, on substituting values in midpoint formula, we get

$(-5,1)=\left(\frac{-4+x_{2}}{2}, \frac{-5+y_{2}}{2}\right)$

On comparing, with the formula,

$\frac{-4+x_{2}}{2}=-5 \text { and } \frac{-5+y_{2}}{2}=1$

$-4+\mathrm{x}_{2}=-10 \text { and }-5+\mathrm{y}_{2}=2$

$\mathrm{x}_{2}=-6 \text { and } \mathrm{y}_{2}=7$

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John and Martha are contemplating having children, but John’s brother has galactosemia (an autosomal recessive disease) and Mart

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<h2>Answer:</h2>

$Probability=\frac{1}{24}$

<h2>Step-by-step explanation:</h2>

As the question states,

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Therefore, the chances for the John to have the disease is = 2/3

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Martha's great-grandmother also had the disease that means her children definitely carried the disease means probability of 1.

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Probability for the child to have disease will be = 1/2

Now, again the child's child (Martha) probability for having the disease is = 1/2.

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